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Suppose we have some topological space $X$ and we somehow forgot about the topology. A friend of ours knows the topology and offers to tell us for any map $X\to Y$ into any topological space $Y$ whether it is continuous or not. As it turns out, we can use this to recover the topology on $X$ the following way:

Let $Z = \{0,1\}$ with topology $\{\varnothing, \{1\}, Z\}$. For a subset $A\subseteq X$ we have a map \begin{align} f_A : X &\longrightarrow Z\\ x &\longmapsto \begin{cases} 1 & \text{if $x\in A$,}\\ 0 &\text{if $x\notin A$.}\end{cases} \end{align} Now $f_A$ is continuous if and only if the preimages of open sets are open, since $f^{-1}(\varnothing)=\varnothing$ and $f^{-1}(Z)=X$ are open in any topology, we know that $f_A$ is continuous if and only if $f^{-1}(\{1\}) = A$ is open in $X$. Thus, given any subset $A$ in $X$ we ask our friend if $f_A$ is continous and we know if $A$ is open or not, so we recovered the topology as $$ \{\, A\subseteq X \mid \text{$f_A\colon X\to Z$ is continuous}\,\}. $$

We conclude that knowing the topology (i.e. the collection of open sets) of $X$ and being able to tell for any map $X\to Y$ if it is continuous are equivalent.

Is it somehow possible to define a topological space as a set $X$ together with some class of maps from $X$ satisfying certain properties so they turn out to be the continuous maps?

One problem here is that thinking of the class $$\{\, f:X\to Y \mid \text{$Y$ a topological space, $f$ continuous}\,\}$$ already implies we know what the topological spaces $Y$ are, so it seems we cannot use this class to define what a topological space is.

Can we do something similar though?

Christoph
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    The space $Z$ you use is known as the Sierpinski space. I like your question. – Ittay Weiss Apr 16 '14 at 10:10
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    The maps should be closed under composition (so we have in fact a category) and the maps from an arbitrary space to $Z$ should be closed under finite min and arbitrary max and include the constant maps? – Hagen von Eitzen Apr 16 '14 at 10:16
  • @HagenvonEitzen Of course we can translates properties of a collection of subsets $U\subseteq X$ to properties of a collection of characteristic functions $X\to{0,1}$ easily. What I was thinking about was something not involving a special codomain like ${0,1}$, so we have a more intrinsic definition of continuity. – Christoph Apr 16 '14 at 10:25
  • You are aware of the notion of an initial topology? You start with a set of maps $f_i: X \rightarrow Y_i$, for a set of topological spaces $Y_i, i \in I$ and a set $X$. The initial topology on $X$ is the smallest topology that makes all $f_i$ continuous. This is how product topologies are defined (using the set of all projections) and the subspace topology (using the one embedding map). But this is not exact, as you demand; the minimality is essential. – Henno Brandsma Apr 16 '14 at 12:33
  • @HennoBrandsma Yes, I'm aware of that, but this doesn't help in the goal of defining what a topology is by characterizing continuity since we already start with topological spaces $Y_i$, so a definition is already set up. – Christoph Apr 16 '14 at 13:50
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    What you propose is, essentially, to study the category of topological spaces as an abstract category. What do you hope to achieve by doing this? – Zhen Lin Apr 16 '14 at 16:35
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    @ZhenLin I'm not really trying to achieve something. I just thought about the different axiomatizations of topological spaces by open sets, closed sets, closure operator, neighborhood systems, ... and asked myself if we can use continuous maps to do this in one way or another. – Christoph Apr 16 '14 at 16:45
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    @Christoph In your example your $Z$ also has a topology so that's quite similar. Otherwise you cannot discuss continuity, right? In your case you seem to want to define the topology simultaneously, for all sets, and don't want to start with special spaces like the Sierpinski space. – Henno Brandsma Apr 16 '14 at 17:17
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    @HennoBrandsma Exactly! – Christoph Apr 16 '14 at 17:20
  • Sorry for my deleted answer. I realized afterwards I was drastically oversimplifying your question. – roo Apr 17 '14 at 01:33

1 Answers1

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A topology space on $X$, is a set of functions $X \to \{False, True\}$, such that.

  • Constant functions are continuous.
  • For any collection of continuous functions $f, g, h \dots$ (possibly infinite), the function $z(x) = f(x) \vee g(x) \vee h(x) \vee \dots$ is continuous.
  • For any two continuous functions $f$ and $g$, $z(x) = f(x) \wedge g(x)$ is continuous.

(Interesting note: If you let functions be potential computations, $False$ represent infinite loops, and $True$ termination, something interesting happens.)

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Christopher King
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